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Let $R$ be a commutative ring with identity and let $f\colon R^{\oplus n}\to R^{\oplus n}$ be a surjective linear mapping. Show that $\det f\in R^\times$.

This question was inspired by A surjective homomorphism between finite free modules of the same rank .

I am curious about what is the proof mentioned in the question .

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Call $e^1,\cdots, e^n$ the canonical basis of $R^n$, call $F$ the matrix that represents $f$ in that basis and call $F^1,\cdots, F^n$ the colums of $F$. Saying that $\operatorname{col}F=R^n$ means that there are $a_{ij}$ such that for all $j$, $\sum_{i=1}^n a_{ij}F^i=e^j$. In other words, that there is an $n\times n$ matrix $A$ such that $FA=I$. Now, by Binet's theorem, $\det(F)\det(A)=\det(FA)=\det I=1$, q.e.d.

Why Binet holds in commutative rings with 1: Consider the polynomial ring $R[X_1,\cdots, X_n]$. There is a natural homomorphism $u:\Bbb Z[X_1,\cdots, X_n]\to R[X_1,\cdots, X_n]$, $u_R(\sum_{J\in\Bbb N^n} a_JX^J)=\sum_{J\in\Bbb N^n}(a_J)_RX^J$, where $n_R:=\underbrace{1+\cdots+1}_{n\text{ times}}$. Also, given a point $p\in R^n$, call $\nu_p:R[X_1,\cdots,X_n]\to R$ the map $\nu_p(f)=f(p)$. Now, Binet for matrices in $R^{n\times n}$ means that given the following polynomial $b\in \Bbb Z[X_{ij},Y_{ij}\,:\, 1\le i,j\le n]$ $$b=\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n\sum_{j=1}^nX_{ij}Y_{j,\sigma(i)}\right)-\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^nX_{i,\sigma(i)}\right)\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^nY_{i,\sigma(i)}\right)$$

we have $\nu_p(u_R(b))=0$ for all $p\in R^{n\times n}\times R^{n\times n}$. However, by the identity principle of polynomials, proving Binet in the special case where $R$ is any infinite domain (for instance, the field $\Bbb R$ or $\Bbb Q$) implies that $b=0$.

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